Victory for the ACT Student Text 15e

460 • C AMBRIDGE P RACTICE T EST R EINFORCEMENT

First, rewrite the

inequality as an equation, factor, and solve: x x 2 3 0

36. (F) (p. 386) Mathematics/Algebra and Functions/Solving Quadratic Equations and Relations .

( 2 − − = − + = = = − Next, plot 1 and 3 on a number line and test each of the three intervals. Select any value from each of the three intervals, and determine if the inequality holds true. ”‘ –Š‡ ϐ‹”•– ‹–‡”˜ƒŽǡ x 1 < − , test x ͷ ‹ –Š‡ ‘”‹‰‹ƒŽ ‹‡“—ƒŽ‹–›ǣ ( ) ( ) 5 2 5 3 0 32 0 > > 2 & − − − − , which is true. This means we can eliminate (H) and (J). Test the values x 3 and x 1 = − to determine whether (F), (G), or (K) is correct: each results in the original inequality being Ͳ Ͳ ! , which is false, so they are not included in the solution set. Note that “not included” in the solution is indicated by an open circle above a value on the number line. 37. (D) (p. 387) Mathematics/Geometry/Trigonometry/Trigonometric Relationships . To simplify this expression, a common denominator is required. The lowest common denominator is sin x : sin sin cos in cos 2 2 2 ) ( 3 1 0 ) x x x x 3 1 or

sin d n Since sin cos x 2 + x x x 1 +

x

x

x

=

+

x

x

x

sin

sin

sin

+

x

x

2

2

sin cos

=

x

sin , the expression reduces to . 38. (K) (p. 387) Mathematics/Number and Quantity/Basic Arithmetic Manipulations . x 1 2 = sin x 1

Perform the indicated operations.

Š‡ ‹ †‘—„–ǡ ™”‹–‡ ‹– ‘—–ǣ . . 2 5 10 5 2 5

^

h

^

h

^

h

=

5

#

#

#

. 2 5 10 2 5 10

^

^ h

= = =

2 5 2

( . ) 2 5 10

h

+

5 5

( . ) ( . ) 2 5 2 5 10

10

#

. 6 25 10

39. (E) (p. 387) Mathematics/Algebra and Functions/Evaluating, Interpreting, and Creating Algebraic Functions/Function Notation . Substitute 2 for x in ( ) f x and evaluate: ( ) 2

− − −

x x x

=

f x

( ) ( ) − − − 2

2 2 2

=

f

( ) 2

2 − − −

4 2

=

− −

6

=

2

=

3